Solve for $x$ : $ 6|x - 4| + 3 = -3|x - 4| + 6 $
Solution: Add $ {3|x - 4|} $ to both sides: $ \begin{eqnarray} 6|x - 4| + 3 &=& -3|x - 4| + 6 \\ \\ { + 3|x - 4|} && { + 3|x - 4|} \\ \\ 9|x - 4| + 3 &=& 6 \end{eqnarray} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} 9|x - 4| + 3 &=& 6 \\ \\ { - 3} &=& { - 3} \\ \\ 9|x - 4| &=& 3 \end{eqnarray} $ Divide both sides by ${9}$ $ \dfrac{9|x - 4|} {{9}} = \dfrac{3} {{9}} $ Simplify: $ |x - 4| = \dfrac{1}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 4 = -\dfrac{1}{3} $ or $ x - 4 = \dfrac{1}{3} $ Solve for the solution where $x - 4$ is negative: $ x - 4 = -\dfrac{1}{3} $ Add ${4}$ to both sides: $ \begin{eqnarray} x - 4 &=& -\dfrac{1}{3} \\ \\ {+ 4} && {+ 4} \\ \\ x &=& -\dfrac{1}{3} + 4 \end{eqnarray} $ Change the ${ + 4}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{1}{3} {+ \dfrac{12}{3}} $ $ x = \dfrac{11}{3} $ Then calculate the solution where $x - 4$ is positive: $ x - 4 = \dfrac{1}{3} $ Add ${4}$ to both sides: $ \begin{eqnarray} x - 4 &=& \dfrac{1}{3} \\ \\ {+ 4} && {+ 4} \\ \\ x &=& \dfrac{1}{3} + 4 \end{eqnarray} $ Change the ${ + 4}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{1}{3} {+ \dfrac{12}{3}} $ $ x = \dfrac{13}{3} $ Thus, the correct answer is $x = \dfrac{11}{3} $ or $x = \dfrac{13}{3} $.